`HA + OH rightarrow H_(2)O + A^(-+) q_(1) kJ
H^(+) + OH ^(-) rightarrow H_(2)O + q_(2)kJ` the enthalpy of ionisation of HA is

Options

A. `(q_(1) + q_(2))`
B. `(q_(1) - q_(2))`
C. `(q_(2) - q_(1))`
D. `-(q_(1) + q_(2))`

Answer

To find the enthalpy of ionization of HA, we can analyze the two given reactions step by step.

### Step 1: Write down the reactions
We have two reactions:
1. \( \text{HA} + \text{OH}^- \rightarrow \text{H}_2\text{O} + \text{A}^- \) with heat change \( q_1 \) kJ.
2. \( \text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O} \) with heat change \( q_2 \) kJ.

### Step 2: Understand the enthalpy of ionization
The enthalpy of ionization of HA can be represented as:
\[ \text{HA} \rightarrow \text{H}^+ + \text{A}^- \]

### Step 3: Manipulate the reactions
To find the enthalpy change for the ionization of HA, we can manipulate the two reactions. We can reverse the second reaction to express it in terms of the ionization of HA.

Reversing the second reaction gives:
\[ \text{H}_2\text{O} \rightarrow \text{H}^+ + \text{OH}^- \quad \text{with heat change} -q_2 \text{ kJ} \]

### Step 4: Add the reactions
Now, we can add the first reaction and the reversed second reaction:
1. \( \text{HA} + \text{OH}^- \rightarrow \text{H}_2\text{O} + \text{A}^- \) (with \( q_1 \))
2. \( \text{H}_2\text{O} \rightarrow \text{H}^+ + \text{OH}^- \) (with \( -q_2 \))

When we add these two reactions, we get:
\[ \text{HA} + \text{OH}^- + \text{H}_2\text{O} \rightarrow \text{H}_2\text{O} + \text{A}^- + \text{H}^+ + \text{OH}^- \]

### Step 5: Cancel common species
Now, we can cancel the common species on both sides:
- \( \text{H}_2\text{O} \) cancels out
- \( \text{OH}^- \) cancels out

This leaves us with:
\[ \text{HA} \rightarrow \text{H}^+ + \text{A}^- \]

### Step 6: Calculate the enthalpy change
The total heat change for this reaction is:
\[ \Delta H = q_1 - q_2 \]

Thus, the enthalpy of ionization of HA is given by:
\[ \Delta H = q_2 - q_1 \]

### Final Answer
The enthalpy of ionization of HA is:
\[ \Delta H = q_2 - q_1 \text{ kJ} \]

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